Find the equation of the parabola with a focus (2, 5) and directrix of y = 3.

A.)
y=x^2/4 + x + 5

B.)
y=x^2/4 + x -5

C.)
y=x^2/4 -x - 5

D.)
y=x^2/4 - x + 5

Respuesta :

Answer: [tex]y=\frac{x^{2}}{4}-x+5[/tex]

Step-by-step explanation:

As the vertex is equidistant from the focus and the directrix, the vertex is (2,4).

This means the equation of the parabola is of the form:

[tex](x-2)^2 = 4p(y-4)[/tex]

The directrix is y = 3, which is the same as y = k - p = y = 4-p, so, so p=1.

So, the equation is:

[tex](x-2)^2 = 4(y-4)\\\\x^2 -4x+4=4(y-4)\\\\ \frac{x^{2}}{4}-x+1=y-4\\\\\boxed{y=\frac{x^{2}}{4}-x+5}[/tex]

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